Integrand size = 27, antiderivative size = 84 \[ \int \cos (c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {1}{2} (A b+a B) x+\frac {(3 a A+2 b B) \sin (c+d x)}{3 d}+\frac {(A b+a B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b B \cos ^2(c+d x) \sin (c+d x)}{3 d} \] Output:
1/2*(A*b+B*a)*x+1/3*(3*A*a+2*B*b)*sin(d*x+c)/d+1/2*(A*b+B*a)*cos(d*x+c)*si n(d*x+c)/d+1/3*b*B*cos(d*x+c)^2*sin(d*x+c)/d
Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.89 \[ \int \cos (c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {6 A b c+6 a B c+6 A b d x+6 a B d x+3 (4 a A+3 b B) \sin (c+d x)+3 (A b+a B) \sin (2 (c+d x))+b B \sin (3 (c+d x))}{12 d} \] Input:
Integrate[Cos[c + d*x]*(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x]),x]
Output:
(6*A*b*c + 6*a*B*c + 6*A*b*d*x + 6*a*B*d*x + 3*(4*a*A + 3*b*B)*Sin[c + d*x ] + 3*(A*b + a*B)*Sin[2*(c + d*x)] + b*B*Sin[3*(c + d*x)])/(12*d)
Time = 0.41 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 3447, 3042, 3502, 3042, 3213}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cos (c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \int \cos (c+d x) \left ((a B+A b) \cos (c+d x)+a A+b B \cos ^2(c+d x)\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left ((a B+A b) \sin \left (c+d x+\frac {\pi }{2}\right )+a A+b B \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {1}{3} \int \cos (c+d x) (3 a A+2 b B+3 (A b+a B) \cos (c+d x))dx+\frac {b B \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (3 a A+2 b B+3 (A b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {b B \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {1}{3} \left (\frac {(3 a A+2 b B) \sin (c+d x)}{d}+\frac {3 (a B+A b) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3}{2} x (a B+A b)\right )+\frac {b B \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
Input:
Int[Cos[c + d*x]*(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x]),x]
Output:
(b*B*Cos[c + d*x]^2*Sin[c + d*x])/(3*d) + ((3*(A*b + a*B)*x)/2 + ((3*a*A + 2*b*B)*Sin[c + d*x])/d + (3*(A*b + a*B)*Cos[c + d*x]*Sin[c + d*x])/(2*d)) /3
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 5.36 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77
method | result | size |
parallelrisch | \(\frac {3 \left (A b +B a \right ) \sin \left (2 d x +2 c \right )+B b \sin \left (3 d x +3 c \right )+3 \left (4 A a +3 B b \right ) \sin \left (d x +c \right )+6 \left (A b +B a \right ) x d}{12 d}\) | \(65\) |
parts | \(\frac {\left (A b +B a \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\sin \left (d x +c \right ) A a}{d}+\frac {B b \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3 d}\) | \(70\) |
derivativedivides | \(\frac {\frac {B b \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+A b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A a \sin \left (d x +c \right )}{d}\) | \(85\) |
default | \(\frac {\frac {B b \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+A b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A a \sin \left (d x +c \right )}{d}\) | \(85\) |
risch | \(\frac {x A b}{2}+\frac {a B x}{2}+\frac {\sin \left (d x +c \right ) A a}{d}+\frac {3 b B \sin \left (d x +c \right )}{4 d}+\frac {B b \sin \left (3 d x +3 c \right )}{12 d}+\frac {\sin \left (2 d x +2 c \right ) A b}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B a}{4 d}\) | \(85\) |
norman | \(\frac {\left (\frac {A b}{2}+\frac {B a}{2}\right ) x +\left (\frac {A b}{2}+\frac {B a}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {3 A b}{2}+\frac {3 B a}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3 A b}{2}+\frac {3 B a}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {\left (2 A a -A b -B a +2 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {\left (2 A a +A b +B a +2 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 \left (3 A a +B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(179\) |
orering | \(\text {Expression too large to display}\) | \(868\) |
Input:
int(cos(d*x+c)*(a+cos(d*x+c)*b)*(A+B*cos(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/12*(3*(A*b+B*a)*sin(2*d*x+2*c)+B*b*sin(3*d*x+3*c)+3*(4*A*a+3*B*b)*sin(d* x+c)+6*(A*b+B*a)*x*d)/d
Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int \cos (c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {3 \, {\left (B a + A b\right )} d x + {\left (2 \, B b \cos \left (d x + c\right )^{2} + 6 \, A a + 4 \, B b + 3 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)*(a+b*cos(d*x+c))*(A+B*cos(d*x+c)),x, algorithm="frica s")
Output:
1/6*(3*(B*a + A*b)*d*x + (2*B*b*cos(d*x + c)^2 + 6*A*a + 4*B*b + 3*(B*a + A*b)*cos(d*x + c))*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (76) = 152\).
Time = 0.14 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.00 \[ \int \cos (c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\begin {cases} \frac {A a \sin {\left (c + d x \right )}}{d} + \frac {A b x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A b x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {B a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 B b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a + b \cos {\left (c \right )}\right ) \cos {\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*(a+b*cos(d*x+c))*(A+B*cos(d*x+c)),x)
Output:
Piecewise((A*a*sin(c + d*x)/d + A*b*x*sin(c + d*x)**2/2 + A*b*x*cos(c + d* x)**2/2 + A*b*sin(c + d*x)*cos(c + d*x)/(2*d) + B*a*x*sin(c + d*x)**2/2 + B*a*x*cos(c + d*x)**2/2 + B*a*sin(c + d*x)*cos(c + d*x)/(2*d) + 2*B*b*sin( c + d*x)**3/(3*d) + B*b*sin(c + d*x)*cos(c + d*x)**2/d, Ne(d, 0)), (x*(A + B*cos(c))*(a + b*cos(c))*cos(c), True))
Time = 0.10 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.94 \[ \int \cos (c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b + 12 \, A a \sin \left (d x + c\right )}{12 \, d} \] Input:
integrate(cos(d*x+c)*(a+b*cos(d*x+c))*(A+B*cos(d*x+c)),x, algorithm="maxim a")
Output:
1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*b - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*b + 12*A*a*sin(d*x + c ))/d
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.81 \[ \int \cos (c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {1}{2} \, {\left (B a + A b\right )} x + \frac {B b \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (B a + A b\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, A a + 3 \, B b\right )} \sin \left (d x + c\right )}{4 \, d} \] Input:
integrate(cos(d*x+c)*(a+b*cos(d*x+c))*(A+B*cos(d*x+c)),x, algorithm="giac" )
Output:
1/2*(B*a + A*b)*x + 1/12*B*b*sin(3*d*x + 3*c)/d + 1/4*(B*a + A*b)*sin(2*d* x + 2*c)/d + 1/4*(4*A*a + 3*B*b)*sin(d*x + c)/d
Time = 41.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {A\,b\,x}{2}+\frac {B\,a\,x}{2}+\frac {A\,a\,\sin \left (c+d\,x\right )}{d}+\frac {3\,B\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,b\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \] Input:
int(cos(c + d*x)*(A + B*cos(c + d*x))*(a + b*cos(c + d*x)),x)
Output:
(A*b*x)/2 + (B*a*x)/2 + (A*a*sin(c + d*x))/d + (3*B*b*sin(c + d*x))/(4*d) + (A*b*sin(2*c + 2*d*x))/(4*d) + (B*a*sin(2*c + 2*d*x))/(4*d) + (B*b*sin(3 *c + 3*d*x))/(12*d)
Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.75 \[ \int \cos (c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -\sin \left (d x +c \right )^{3} b^{2}+3 \sin \left (d x +c \right ) a^{2}+3 \sin \left (d x +c \right ) b^{2}+3 a b d x}{3 d} \] Input:
int(cos(d*x+c)*(a+b*cos(d*x+c))*(A+B*cos(d*x+c)),x)
Output:
(3*cos(c + d*x)*sin(c + d*x)*a*b - sin(c + d*x)**3*b**2 + 3*sin(c + d*x)*a **2 + 3*sin(c + d*x)*b**2 + 3*a*b*d*x)/(3*d)